Biostat 202B Homework 1

Due April 11, 2024 @ 11:59PM

Question 1

(a) Answer:

Let \(g(x)=\sqrt{x}\), g is continuous for \(x\geq 0\).

Since \(P(X_i >0)=1\) for all i, all \(X_i\) are greater than 0.

Since \(X_n\rightarrow a\) in probability,

\(Y_n = \sqrt{X_n} = g(X_n) \rightarrow \sqrt{a}\) in probability by C.M.T

Let \(q(x)=\frac{a}{x}\), q is continuous for \(x > 0\).

Since \(X_n\rightarrow a\) in probability,

\(Y_n' = q(X_n)=\frac{a}{X_n}\rightarrow \frac{a}{a}=1\) in probability by C.M.T

(b) Answer:

Since \(S_n^2\rightarrow \sigma^2\) in probability

\(S_n = \sqrt{S_n^2}\)

Hence, by part (a)

\(S_n\rightarrow \sqrt{\sigma^2}=\sigma\) in probability

\(\frac{\sigma}{S_n}\rightarrow \frac{\sigma}{\sigma}=1\) in probability

Question 2

(a) Answer:

Since \(Var(X_n)\rightarrow 0\) in probability

\((X_n-E(X_n))^2\rightarrow 0\) in probability

\(X_n-E(X_n)\rightarrow 0\) in probability

Since \(E(X_n)\rightarrow \mu\) in probability

\(X_n\rightarrow \mu\) in probability for \(n\rightarrow \infty\)

Question 3

Answer:

\(\mu = E(X)=\int_{-1}^{1}x\frac{1}{2}(1+\theta x)dx=(\frac{x^2}{4}+\frac{\theta x^3}{6})\Big|_{-1}^{1}=\frac{2\theta}{6}\)

Then, \(\theta = 3\mu\)

Since \(\bar{X}_n\rightarrow \mu\) in probability by WLLN,

\(3\bar{X}_n\rightarrow 3\mu=\theta\) in probability

Hence, \(3\bar{X}_n\) is a consistent estimator of \(\theta\)

Question 4

(a) Answer:

\(E(Y_n)=np\) and \(Var(Y_n)=np(1-p)\)

\(E(\frac{Y_n}{n})=\frac{np}{n}=p\) and \(Var(\frac{Y_n}{n})=\frac{np(1-p)}{n^2}=\frac{p(1-p)}{n}\)

By Chebyshev inequality:

\[\begin{align} P(|\frac{Y_n}{n}-p|>\epsilon)&=P(|\frac{Y_n}{n}-E(\frac{Y_n}{n})|>\epsilon) \\ &\leq\frac{Var(\frac{Y_n}{n})}{\epsilon^2}\\ &=\frac{p(1-p)}{n\epsilon^2}\rightarrow 0 \text{, as n goes large} \end{align}\]

Since \(P(|\frac{Y_n}{n}-p|>\epsilon)\geq 0\),

\(P(|\frac{Y_n}{n}-p|>\epsilon)\rightarrow 0\) in probability because it is squeezed on both sides of inequalities.

Hence, \(\frac{Y_n}{n}\rightarrow p\) in probability.

(b) Answer:

\(Var(1-\frac{Y_n}{n})=\frac{p(1-p)}{n}\)

\[\begin{align} P(|1-\frac{Y_n}{n}-(1-p)|<\epsilon)&=P(|1-\frac{Y_n}{n}-E(1-\frac{Y_n}{n})|<\epsilon)\\ &\geq \frac{Var(1-\frac{Y_n}{n})}{\epsilon^2}\\ &=\frac{p(1-p)}{n \epsilon^2}\rightarrow 0 \text{ in probability} \end{align}\]

Since \(P(|1-\frac{Y_n}{n}-(1-p)|<\epsilon)\geq 0\),

Then, \(1-\frac{Y_n}{n}\rightarrow 1-p\) in probability.

(c) Answer:

Let \(g(x)=x(1-x)\) be continuous for all x

then, \(\frac{Y_n}{n}(1-\frac{Y_n}{n})=g(\frac{Y_n}{n})= g(p)\rightarrow p(1-p)\) in probability by C.M.T

Question 5

Answer:

Since p>0,

\(P(|W_n-\mu|>\epsilon)=P(|W_n-E(W_n)|>\epsilon)\leq \frac{Var(W_n)}{\epsilon^2}=\frac{b}{n^p\epsilon^2}\rightarrow 0\) in probability

Since \(P(|W_n-\mu|>\epsilon)\geq 0\)

\(P(|W_n-\mu|>\epsilon)\rightarrow 0\) in probability because it is squeezed on both sides

Hence, \(W_n\rightarrow \mu\) in probability